Answer:
The appropriate response will be "Length must be increased by 0.012%".
Explanation:
The given values is:
ΔT = 5 s/day
Now,
⇒ 
On multiplying both sides by "100", we get
⇒ 
⇒ 
(%)
∵ 
On substituting the values, we get
⇒ 
% = 
%
On applying cross multiplication, we get
⇒ 
% = 
%
⇒ = 
⇒ = 
⇒ = 
%
Answer:
48.51ms / 174.6 km/h
Explanation:
y = 1/2 x g x t^2 v = g x t
when y = 120m
120 = 1/2 x 9.8 x t^2
t^2 = 24.49
t = 4.95s
when t = 4.95s
v = 9.8 x 4.95
v = 48.51 m/s = 174.6 km/h
I'd say its realistic. But I don't really know that sry
It would be helpful if you gave me a bit more information on what the cars speed is
Answer:
a) w = 7.27 * 10^-5 rad/s
b) v1 = 463.1 m/s
c) v1 = 440.433 m/s
Explanation:
Given:-
- The radius of the earth, R = 6.37 * 10 ^6 m
- The time period for 1 revolution T = 24 hrs
Find:
What is the earth's angular speed?
What is the speed of a point on the equator?
What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?
Solution:
- The angular speed w of the earth can be related with the Time period T of the earth revolution by:
w = 2π / T
w = 2π / 24*3600
w = 7.27 * 10^-5 rad/s
- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.
v1 = R*w
v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)
v1 = 463.1 m/s
- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.
π/2 ........... s
x ............ 1/5 s
x = π/2*5 = 18°
- The radius of the earth R' at point where θ = 18° from the equator is:
R' = R*cos(18)
R' = (6.37 * 10 ^6)*cos(18)
R' = 6058230.0088 m
- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.
v2 = R'*w
v2 = (6058230.0088)*(7.27 * 10^-5)
v2 = 440.433 m/s
The correct answer is C , because the space is vacuum and his body can explode and for this reason, the astronaut need a special costum to be protected. It's the same on the moon, because there is no atmosphere
